Point of Inflection

Date: 05/17/97 at 12:47:07
From: Andrew Reniers
Subject: A Calculus Question
Hi Dr. Math,
I have this question that I have tried and don't understand. I have
no clue what to do. I hope that you can help me. Here it is:
Suppose that the cubic polynomial h(x) = x^3 - 3bx^2 + 3cx + d has a
local maximum at A: (x1, y1) and a local minimum at B: (x2, y2).
Prove that the point of inflection of h is at the midpoint of the line
segment AB.
I hope you can help me with this problem.
Thank you,
Andrew Reniers
P.S.: This web site is a great idea!

Date: 05/19/97 at 05:45:37
From: Doctor Anthony
Subject: Re: A Calculus Question
It is easy to show that the x-coordinate of the point of inflection is
midway between the maximum and minimum turning points.
h'(x) = 3x^2 - 6bx + 3c = 0 for max and min turning points.
x^2 - 2bx + c = 0
The sum of the roots is 2b, so we have x1+x2 = 2b
(x1+x2)/2 = b
Also the product of the roots is equal to c, so x1*x2 = c
h''(x) = 6x - 6b = 0 for point of inflection
x - b = 0
x = b at point of inflection = (x1+x2)/2
So the x-coordinate of point of inflection is midway between the
maximum and minimum turning points.
We must now consider the y-coordinate of the point of inflection.
We have y1 = x1^3 - 3b.x1^2 + 3c.x1 + d
y2 = x2^3 - 3b.x2^2 + 3c.x2 + d
(y1+y2)/2 = (x1^3+x2^3)/2 - 3b(x1^2+x2^2)/2 + 3c(x1+x2)/2 + d
Now put b = (x1+x2)/2 and c = x1*x2
This simplifies to:
(y1+y2)/2 = -x1^3/4 -x2^3/4 + 3x1^2.x2/4 + 3x1.x2^2/4 + d (1)
Finally put x = (x1+x2)/2 into the equation of the cubic, and simplify
again with b = (x1+x2)/2 and c = x1*x2 and we will get the same
expression as (1) above. It follows that the point
[(x1+x2)/2, (y1+y2)/2] is a point on the curve. We already know that
the point of inflection is at x = (x1+x2)/2 and so the y-value of the
point of inflection is (y1+y2)/2.
So the point of inflection is at the midpoint of the line joining
(x1,y1) and (x2,y2).
-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/